Answer:
Assuming that O₂ is the limiting reagent, approximately 1.6 liters of NO will be produced.
Explanation:
Make sure that the equation has been balanced.
The coefficient in front of [tex]\rm NO[/tex] is [tex]4[/tex]; the coefficient in front of [tex]\rm O_2[/tex] is [tex]5[/tex]. The ratio between these two coefficients is equal to [tex]\displaystyle \frac{4}{5} = 0.8[/tex].
Assume that [tex]\rm O_2[/tex] is indeed the limiting reagent. In other words, assume that [tex]\rm O_2[/tex] runs out before all [tex]\rm NH_3[/tex] is consumed. The ratio between the number of moles of [tex]\rm NO[/tex] produced and the number of moles of [tex]\rm O_2[/tex] consumed will also be equal to [tex]0.8[/tex]. That is:
[tex]\displaystyle \frac{n(\mathrm{NO})}{n(\mathrm{O_2})} = \frac{\text{Coefficient of $\mathrm{NO}$}}{\text{Coefficient of $\mathrm{O_2}$}} = 0.8[/tex].
Additionally, both [tex]\rm NO[/tex] and [tex]\rm O_2[/tex] are gases. Under the same temperature and pressure, one mole of each gas would occupy about the same volume. As a result,
[tex]\displaystyle \frac{V(\mathrm{NO})}{V(\mathrm{O_2})} = \frac{n(\mathrm{NO})}{n(\mathrm{O_2})} = 0.8[/tex].
The volume of [tex]\rm O_2[/tex] is already known. Hence, the volume of [tex]\rm NO[/tex] produced will be equal to:
[tex]\begin{aligned} V(\mathrm{NO}) &= \frac{V(\mathrm{NO})}{V(\mathrm{O_2})} \cdot V(\mathrm{O_2}) \cr &= 0.8 \times (2.0\; \rm L) \cr &= \rm 1.6\; L\end{aligned}[/tex].
